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Find the sum of the series:

  • $ \displaystyle \cdots + \frac{1}{z^{3}} + \frac{1}{z^{2}} + \frac{1}{z} + 1 + z + z^{2} + z^{3} \cdots$

This series can be summed in the following way:

$$\cdots + \frac{1}{z^{3}} + \frac{1}{z^{2}} + \frac{1}{z} + 1 = \frac{z}{z-1}$$ and $$ z + z^{2} + z^{3} + \cdots = \frac{z}{1-z}$$

So the sum equals $0$.

Is this correct or wrong? Please let me know if there is any error.

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I don't understand the -1. This is a genuine question people might have. – Aryabhata Jan 27 '11 at 15:35
@Moron: Perhaps, people might think that as a graduate student i should have done this. – anonymous Jan 27 '11 at 16:01
@Chandru: People should (usually) be judging the content, not the poster. But I guess that could be one reason... – Aryabhata Jan 27 '11 at 16:02
up vote 8 down vote accepted

The error is the following: Your first summation only holds for $\lvert z\rvert >1$ while the second one only holds for $\lvert z\rvert <1$.

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Nice observation. – anonymous Jan 27 '11 at 15:06
@Chandru1: But you knew that already, didn't you? So why did you ask this question? – TonyK Jan 27 '11 at 17:42
@TonyK: What did i know? Look, don't conclude anything yourself. – anonymous Jan 27 '11 at 17:50

This series does not converge for any complex value of $z$. However, in a certain sense if you set $z = e^{2 \pi i t}$, this is the Fourier series of the Dirac delta function.

In any case, you can show that if there is a consistent way to interpret the sum, then for $z \neq 1$ it must be $0$. This is because the expression $f$ satisfies $f = zf$.

(It is possible to work abstractly with two-sided Laurent series, but you cannot multiply them, and you cannot evaluate them anywhere; the best you can do is consider them as a module over the Laurent polynomials. This is done, for example, in certain branches of mathematical physics.)

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